Question

according to the transportation security administration (tsa) data, 6.3% of the passengers fly first - class. a random sample of size 176 was obtained. let $widehat{p}$ be the proportion of the sample that fly first - class. 1. use the central limit theorem (select an answer) to describe the probability distribution of $widehat{p}$ and state its parameters $mu_{widehat{p}}$ and $sigma_{widehat{p}}$ (4 decimal places) $widehat{p}sim$ np $leq$ 10 and n(1 - p) $leq$ 10, np $geq$ 10 and n(1 - p) $geq$ 10, n(1 - p) $leq$ 10 but np $geq$ 10, n(1 - p) $leq$ 10 but np $geq$ 10. 2. find the probability that between.826 (round the answer to 4 decimal places) fly first - class. > next question

Explanation:

Step1: Check Central Limit Theorem conditions

Given $p = 0.063$ and $n=176$. Calculate $np=176\times0.063 = 11.088\geq10$ and $n(1 - p)=176\times(1 - 0.063)=176\times0.937 = 164.812\geq10$. So the condition for normal - approximation of the sampling distribution of $\hat{p}$ is $np\geq10$ and $n(1 - p)\geq10$.

Step2: Calculate the mean of the sampling distribution of $\hat{p}$

The mean of the sampling distribution of the sample proportion $\hat{p}$ is $\mu_{\hat{p}}=p$. So $\mu_{\hat{p}}=0.0630$.

Step3: Calculate the standard deviation of the sampling distribution of $\hat{p}$

The standard deviation of the sampling distribution of the sample proportion is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. Substitute $p = 0.063$ and $n = 176$ into the formula: $\sigma_{\hat{p}}=\sqrt{\frac{0.063\times(1 - 0.063)}{176}}=\sqrt{\frac{0.063\times0.937}{176}}=\sqrt{\frac{0.059031}{176}}\approx0.0183$.

Answer:

1. $np\geq10$ and $n(1 - p)\geq10$, $\mu_{\hat{p}} = 0.0630$, $\sigma_{\hat{p}}=0.0183$
2. (No full information for part 2 in the question, unable to complete this part. If we assume the full - question for part 2 is to find $P(a<\hat{p}